At the equator a 1000 turn coil with a cross sectional area of 300 cm2 and a resistance of 15 Omega is aligned with its plane perpendicular to the earth’s magnetic field of 7*10^-5 T . If the coil is flipped over how much charge flows through it?

May 23, 2018

Important information missing: Time taken to flip over the coil
Let this be $= t$. However, we will see that it is not required for the final answer.

We know that the average induced current ${I}_{a v}$ in the coil is given by Faraday’s law of induction.

${I}_{a v} = \frac{\varepsilon}{R}$ .........(1)
where $\varepsilon$ is the induced emf produced due to change in flux linked with the coil and $R$ is the resistance of coil.

We also know from Lenz's law that the induced emf in the coil is the rate at which the magnetic flux linked with the coil changes and opposes it.

$\varepsilon = - \frac{\Delta \phi}{t}$ .......(2)

The coil of $N$ turns of area $A$ has flux $\phi$ linked with it when it is initially aligned with its plane perpendicular to the earth’s magnetic field $B$. As the coil is flipped over the linked flux becomes $- \phi$. As such

$\Delta \phi = - 2 \phi = - 2 N B A$ ......(3)

Rewriting (1) with the helps of (2) and (3) we get

${I}_{a v} = \frac{2 N B A}{R t}$ ......(4)

Now charge through coil in time $t$ can be calculated from the definition of current.

$I = \frac{Q}{t}$
$\implies Q = I t$ ......(5)

Using (4) we get

$Q = \frac{2 N B A}{R t} \times t$
$\implies Q = \frac{2 N B A}{R}$

Inserting given values in SI units we get

$Q = \frac{2 \times 1000 \times \left(7 \times {10}^{-} 5\right) \times \frac{300}{10} ^ 4}{15}$
$\implies Q = 2.8 \times {10}^{-} 4 \setminus C$